Subspace operations in affine
Klingenberg spaces
T. Bisztriczky
J.W. (Michael) Lorimer
In two previous papers we introduced the notion of an Affine Klingenberg space
A and presented a geometric description of its free subspaces. Presently, we consider
the operations of join, intersection and parallelism on the free subspaces of A.
As in the case of ordinary affine spaces, we obtain the Parallel Postulate. The
situation with join and intersection is not that straightforward. In particular, the
central problem is whether the join of two free subspaces is free?
We show that if A is not an ordinary affine space and dim A ≥ 4 then A has a
subspace which is both not free and the join of two free subspaces. Thus, join and
intersection do not possess the usual closure properties. We determine necessary
and sufficient conditions under which the join of two free subspaces is free, and in
such a case we verify the Dimension Formula.
The subspace operations are essential tools for establishing when A is desarguesian and when it can be embedded in a projective Klingenberg space.
1
Preliminaries
Let P = {P, Q, ...} be a set of points, L = {`, m, ...} be a set of lines and A =
{P, L, I, k} be an incidence structure with parallelism; that is, k is an equivalence
relation on L such that for each (P, `) ∈ P × L, there is a unique line L(P, `) with
P I L(P, `)k` .
We call A an affine Klingenberg space (AK-space) if there is an equivalence
relation ∼ on P (neighbour relation) such that A = < P, L, I, k, ∼> satisfies the
axioms (A1) to (A7) below:
Received by the editors February 1994
Communicated by J.Thas
AMS Mathematics Subject Classification : 51C05, 51D10, 51D15.
Keywords : affine Klingenberg spaces, subspaces.
Bull. Belg. Math. Soc. 2 (1995), 99–108
100
T. Bisztriczky – J.W. Lorimer
(A1) Any P 6∼ Q are incident with a unique line P ∨ Q, and any line is incident
with at least two non-neighbouring points.
We note from [1] that (A1) permits us to assume that the incidence is inclusion
and so, lines are subsets of P.
A subset Q ⊆ P is a subspace of A if ` ∪ {P, Q} ⊆ Q and P 6∼ Q imply that
P ∨ Q ⊆ Q and L(P, `) ⊆ Q. Clearly, any point and any line is a subspace
of A and s(A), the set of all subpaces of A, is closed under intersections. Let
{Q, R} ⊂ s(A). Then Q is a neighbour of R(Q ≈ R) if each point of Q is a
neighbour to some point of R.
Let Π ⊆ P. Then < Π > : = ∩Q, Π ⊆ Q ∈ s(A), is the subspace generated
by Π. Next, Π is independent if P 6≈ < Π\{P } > for each P ∈ Π, otherwise,
Π is dependent. Finally, Π is a basis of a subspace Q if Π is independent and
Q = < Π > . If Π is a basis of Q ∈ s(A) then Q is called a free subspace of A.
Let f(A) be the set of all free subspaces of A. Then Q ∈ f(A) is a plane if
it has a basis of cardinality three, and H ∈ f(A) is a hyperplane if there is a
maximal independent subset {Pλ }Λ of A such that
H = < {Pλ }λ∈Λ\{λ0 } > for some λ0 ∈ Λ.
We note from [1] that maximal independent subsets of A exist.
(A2) If H is a hyperplane of A, ` ∈ L and ` 6≈ H then |` ∩ H| ≤ 1.
(A3) If {P, Q, R} ⊂ P and P ∼ Q 6∼ R then P ∨ R ≈ Q ∨ R.
We recall that for {p, q, r} ⊂ L, (p, q 6≈ r) means that p, q and r are distinct
and mutually intersecting, q 6≈ r and there exist Q ∈ r ∩ p and R ∈ q ∩ p
such that Q ∈
/ q and R ∈
/ r.
(A4) If {p, p0 , q, q 0, r, r0} ⊂ L such that pkp0 , qkq 0, rkr0 , (p, q 6≈ r) and q 0 ∩ p0 6= ∅ 6=
r0 ∩ p0 then q 0 6≈ r0 and q 0 ∩ r0 6= ∅.
(A5) P contains an independent set of cardinality three.
(A6) Every line contains three mutually non-neighbouring points.
For Π ⊂ P, we call Π := {P ∈ P|P ∼ X for some X ∈ Π}, the saturate of Π.
(A7) If {Q, R} ⊂ f(A) and Q ⊆ R ⊆ Q then Q = R.
Henceforth, let A = < P, L, ∈, k, ∼> be a fixed AK-space. We list some essential
properties of A.
1.1 ([1], 3.3) ≈ is an equivalence relation of L. (We write ∼ for ≈ restricted
to L.)
1.2 ([1], 3.7) Let pkq. Then p ∼ q if and only if P ∼ Q for some P ∈ p and Q ∈ q.
For points P1 , ..., Pn , we set {Pt }0 = {P1 , ..., Pn }, {P̂i }0 = {Pt }0 \{Pi },
n
n
n
n
< Pt >0 =< {Pt }0 > and < P̂i >0 =< {P̂i }0 > .
n
n
n
Subspace operations in affine Klingenberg spaces
101
1.3 ([1],3.13) Let{Pt }0 be independent, n ≥ 1. Then < Pt >0 has the following
properties:
n
n
(B)n For each X ∈< Pt >0 , L(X, Pi ∨ Pj ) intersects < P̂i >0 and < P̂j >0 ,
0 ≤ i 6= j ≤ n.
n
n
n
(E)n If {Qt }0 is independent and Pt ∼ Qt for 0 ≤ t ≤ n then
n
< Pt >0 ≈ < Qt >0 ≈ < Pt >0 .
n
n
n
(F )n−1 If Q 6≈ < P̂i >0 then {P̂i }0 ∪ {Q} is independent.
n
n
(I)n If {Qt }0 ⊆< Pt >0 is independent then m ≤ n and there are Qm+1 , ..., Qn
n
n
such that < Qt >0 =< Pt >0 .
m
n
1.4 ([1],3.15 and 3.17) Let A∗ =< P∗ , L∗ , I ∗, k∗ > be the incidence structure with
parallelism where P ∗ : = {Q ∈ P|Q ∼ P }, `∗ : = {m ∈ L{m ≈ `} and
P∗ : = P/ ∼ = {P ∗ |P ∈ P},
L∗ : = L/ ∼ = {`∗ ∈ L},
P ∗ I ∗`∗ ⇐⇒ there exist Q ∼ P and m ∼ ` such that P ∈ m and Q ∈ ` ,
`∗ k∗ m∗ ⇐⇒ there exist `1 ∼ ` and m1 ∼ m such that `1 km1 .
Then
a) A∗ is an affine space,
b) ∗ : A → A∗ is an incidence preserving epimorphism,
c) if {Pλ }Λ is independent then < Pλ >∗Λ = < Pλ∗ >Λ ,
d) {Pλ }Λ is independent if and only if the following two conditions hold: (i)
{Pλ∗ }Λ is independent and (ii) Pα 6= Pβ implies Pα∗ 6= Pβ∗ .
e) the cardinality of every maximal independent subset of A with one point
removed is equal to the dimension of A∗.
We call A∗, the underlying affine space of A. We note that P ∗ = Q∗ if and
only if P = Q, and Π∗ = (Π)∗ for any Π ⊆ P.
For {Q, R} ⊂ s(A), we set Q ∨ R =< Q ∪ R > and call it the join of Q and R.
1.5 ([2],1.8) Let {Pλ }Λ be independent and X, Y 6≈ < Pλ >Λ . Then {X} ∪ {Pλ }Λ
and {Y } ∪ {Pλ }Λ are independent, and if Y ∈ X∨ < Pλ >Λ then Y ∨ < Pλ >Λ
= X∨ < Pλ >Λ .
In [1] and [2], we determined that all maximal independent subsets of Q ∈ s(A)
have the same cardinality. Accordingly, the dimension of Q, dim(Q), is the
cardinality of a maximal independent subset of Q with one point removed.
1.6 ([2],1.11) If Q ∈ s(A) then Q ∈ s(A) and dim(Q) = dim(Q).
1.7 ([2],1.12) Let P ∈ ` . Then |` ∩ P | is independent of the choice of P and ` .
(We call d(A) = |` ∩ P |, the degree of A.)
1.8 ([2].2.5) Let R ∈ f(A).
102
T. Bisztriczky – J.W. Lorimer
a) If R contains a plane then it is an AK-space with the induced parallel
and neighbour relations.
b) If ` 6≈ R then |` ∩ R| ≤ 1.
We set fn (A) = {Q ∈ f(A)| dim(Q) = n}, n ≥ 0, and observe that the underlying affine space A∗, with equality as the neighbour relation on P∗ , is also an
AK-space.
1.9 Proposition. If Q ∈ fn (A) then Q∗ ∈ fn (A∗).
Proof. Let {Qt }0 be a basis of Q. Then Q∗ = (< Qt >0 )∗ =< Q∗t >0 by 1.4 c),
n
and {Q∗t }0 is independent by 1.4 d). Since A∗ is an affine space, it and its subspaces
are free. Thus Q∗ ∈ fn (A∗ ).
0
0
0 0
0
0
A partial AK-space is an incidence structure A =< P , L , I , k , ∼ > with
parallelism k0 and neighbour relation ∼0 which satisfies axioms (A1) to (A6). In
[1] and [2], we called an AK-space a partial AK-space with a weakened axiom (A2)
where a hyperplane H was replaced by a line h. We note that all the results preceding
1.8 are valid in such a weakened partial AK-space, and that 1.8 and the results in
the next section are valid in a partial AK-space A0.
The significance of (A7), as 1.10 below shows, is to ensure that any free subspace
is generated by any of its maximal independent subsets.
1.10 ([2],2.4). Let A0 be a partial AK-space. Then the following statements are
equivalent.
(a) A0 satisfies (A7); that is, A0 is an AK-space.
(b) Every maximal independent subset of a free subspace R is a basis of R.
(c) If H is a hyperplane of a free subspace R, X ∈ R and X 6≈ H then R = H∨X.
Let A0 be a partial AK-space. If A0 has finite dimension n ≥ 2 then (I)n (see 1.3)
and 1.10 imply that A0 is an AK-space. However, if A0 has infinite dimension then
A0 need not be an AK-space. Indeed, utilizing a module described by A. Kreuzer in
[4], we can construct an infinite dimensional A0 , via the procedure delineated in [2],
whose every infinite dimensional free subspace possesses a non-generating maximal
independent subset.
n
2
n
n
Parallelism
Let {Q, R} ⊂ fn (A), n ≥ 2. Then Q and R are parallel (QkR) if there exist bases,
n
n
{Qt }0 of Q and {Rt }0 of R, such that Q0 ∨ Qi kR0 ∨ Ri for i = 1, ..., n.
Our aim is to show that this definition of parallelism for free n-spaces satisfies
the Parallel Postulate. As a first step, we note some results from [2].
2.1 ([2],1.15 and 1.16) Let {Q, R} ⊂ fn (A) and QkR, n ≥ 1.
a) For each line q ⊆ Q, there is a line r ⊆ R such that qkr.
b) Let Q ∈ Q, X 6≈ Q and `kQ ∨ X. Then ` 6≈ R.
Subspace operations in affine Klingenberg spaces
103
2.2 Proposition. Let {Qt }0 be independent, n ≥ 2. For any point R0 , there exist
n
points Rj ∈ rj = L(R0 , Q0 ∨ Qj ), 1 ≤ j ≤ n, such that {Rt }0 is independent and
n
n
Ri ∨ Rj kQi ∨ Qj for 0 ≤ i 6= j ≤ n. In particular, < Qt >0 k < Rt >0 .
n
Proof. Let qij = Qi ∨ Qj for 0 ≤ i 6= j ≤ n.
Let n = 2. Then < Q0 , Q1 , Q2 > is a plane and the qij are mutually nonneighbouring. By (A1), there is a point R1 ∈ r1 , such that R1 6∼ R0 . Next, (A4)
yields that r2 ∩ L(R1 , q12 ) is a point R2 such that R2 ≈ R0 ∨ R1 . Then {Rt }20 is
independent by 1.5, and Ri ∨ Rj kqij for 0 ≤ i 6= j ≤ 2.
Let n ≥ 3 and proceed by induction. Thus there exist points Rj ∈ rj , i ≤ j ≤
n−1
n − 1, such that {Rt }0 is independent and
(1) Ri ∨ Rj kqij for 0 ≤ i 6= j ≤ n − 1.
Since < Q0 , Qj , Qn > is a plane for 1 ≤ j ≤ n − 1, it follows from the preceding
j
j
that rn ∩ L(Rj , qjn ) is a point Rn such that {R0 , Rj , Rn } is independent and
j
j
(2) R0 ∨ Rn kq0n and Rj ∨ Rn kqjn for 1 ≤ j ≤ n − 1.
1
j
We claim that Rn = Rn for j = 2, ..., n − 1. Since < Q1 , Qj , Qn > is a plane,
1
j
(qij , q1n 6∼ qjn ) and q1j kR1 ∨Rj , it follows from (2) and (A4) that (R1 ∨Rn )∩(Rj ∨Rn )
is a point Y such that (R1 , Rj , Y ) is independent. We observe that < Q1 , Qj , Qn >
k < R1 , Rj , Y > and Q0 6≈ < Q1 , Qj , Qn > . Thus rn kQ0 ∨ Qn and 2.1 b) yield
that rn 6≈ < R1 , Rj , Y >, and 1.8 b) implies that |rn ∩ < R1 , Rj , Y > | ≤ 1. Since
1
j
1
j
{Rn , Rn } ⊆ rn ∩ < R1 , Rj , Y >, Rn = Rn .
1
Let Rn = Rn . Then Ri ∨ Rj kqij for 0 ≤ i 6= j ≤ n by (1) and (2) and we need
is independent,
only to show that {Rt}n0 is independent. We recall that {Rt }n−1
0
n−1
n−1
< Qt >0 k < Rt >0 and rn kQ0 ∨ Qn . Hence, as above, rn 6≈ < Rt >n−1
.
0
n−1
Since rn = R0 ∨ Rn and R0 ∈< Rt >0 , it follows readily from 1.4 c) and d) that
. Thus {Rt }n0 is independent by 1.5.
Rn 6≈ < Rt >n−1
0
2.3 Proposition. Let {Q, R} ⊂ fn (A), n ≥ 1. Then QkR if and only if for each
line q ⊆ Q, there is a line r ⊆ R such that qkr.
Proof. The necessity follows from 2.1 a). For the sufficiency, let {Qt }0 be a basis
for Q and choose a point R0 ∈ R. Then for j = 1, ..., n, rj = L(R0 , Q0 ∨ Qj ) ⊆ R.
n
n
By 2.2, there is a point Rj ∈ rj such that {Rt }0 is independent and < Rt >0 kQ.
n
By (I)n , R =< Rt >0 .
n
2.4 Proposition. Parallelism is an equivalence relation on fn (A), n ≥ 1.
Proof. The reflexivity and symmetry follow from the definition. The transitivity
follows from 1.1 and 2.3.
2.5 The Parallel Postulate for fn (A), n ≥ 1. Let Q ∈ fn (A) and R ∈ P. Then
there exists a unique R ∈ fn (A) such that R ∈ RkQ.
Proof. As the assertion is true by assumption for n = 1, let n ≥ 2. Then the
existence of an R ∈ fn (A) such that R ∈ RkQ follows from 2.2.
n
Let {R, R0} ⊂ fn (A) such that R ∈ R ∩ R0 and RkR0 . Let R = R0 and {Rt }0
be a basis of R. Then R0 ∈ R0 , RkR0 and 2.3 yield that L(R0 , R0 ∨ Rj ) ⊂ R0 for
n
j = 1, ..., n. Since R0 ∨ Rj = L(R0 , R0 ∨ Rj ), we have that R0 =< Rt >0 = R by
(I)n .
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T. Bisztriczky – J.W. Lorimer
2.5.1 Corollary. Let Q and R be parallel free n-spaces, n ≥ 1. Then either Q = R
or Q ∩ R = ∅.
2.6 Proposition. Let Q and R be parallel free n-spaces, n ≥ 1. Then Q ≈ R if
and only if Q ∼ R for some Q ∈ Q and R ∈ R.
Proof. As the assertion is true for n = 1 by 1.2, we assume that n ≥ 2 and
proceed by induction. Clearly, we need to verify only the sufficiency.
Let Q0 ∈ Q, R0 ∈ R and Q0 ∼ R0 . By (I)n and QkR, there exist bases, {Qt }n0
n
of Q and {Rt }0 of R, such that Q0 ∨ Qj kR0 ∨ Rj for j = 1, ..., n. For each such j,
< Q̂j >0 k < R̂j >0 and the induction hypothesis yield that < Q̂j >0 ≈ < R̂j >0 .
n
n
n
n
Let Q ∈ Q. By (B)n , L(Q, Q0 ∨ Q1 ) intersects < Q̂1 >0 at a point X. As X ∼ Y
n
for some Y ∈< R̂1 >0 , it follows from 1.2 that
n
L(Q, Q0 ∨ Q1 ) = L(X, Q0 ∨ Q1 ) ∼ L(Y, Q0 ∨ Q1 ) = L(Y, R0 ∨ R1 ).
Since R is a subspace, L(Y , R0 ∨ R1 ) ⊂ R. Thus there is a point R ∈ R such
that Q ∼ R and Q ≈ R.
Finally, we extend parallelism to s(A). Let {C, F} ⊂ s(A). Then C is parallel to
F(CkF) if for any line s ⊆ C, there is a line u ⊆ F such that s||u. We note also that
if Q fm (A), R ∈ fn (A), m < n, and Q||R then it readily follows that Q is parallel
to some free m-space in R.
3
Free meets, free joins, and the dimension formula.
We recall that if the underlying affine space A∗ of A has dimension at least three
then A∗ and all of its subspaces are free, it is desarguesian and can be coordinatized
by a vector space over a skew field. Let Q∗ and R∗ be finite dimensional subspaces
of A∗ and Q∗ ∩ R∗ 6= ∅. Then Q∗ ∨ R∗ is finite dimensional and with vector space
techniques (cf. [3], pp. 15-19), one obtains the dimension formula:
(DF) dim(Q∗ ∨ R∗) + dim(Q∗ ∩ R∗ ) = dim(Q∗ ) + dim(R∗ ).
With regard to the AK-space A , we do not know if A is free or even if it is
desarguesian when it has dimension at least three. Hence, before considering any
generalization of (DF), we examine when Q ∈ fm (A) and R ∈ fn (A) imply that
Q ∨ R and Q ∩ R are free.
First, we need to assume that Q 6≈ R. For if Q 6⊆ R and Q ≈ R then
R 6= Q ∨ R, R ⊆ Q ∨ R ⊆ R and (A7) yield that Q ∨ R is not free.
Second, it is not sufficient to assume only that Q 6≈ R and Q ∩ R 6= ∅.
3.1 Example. Let A be an AK-space with dimension at least four and degree at
least two.
Let {P , Q1, R1 , R2 , S} ⊆ P be independent.
Then F = hP , Q1, R1 , R2 i f3 (A) and S 6≈ F. Let ` = R2 ∨ S. Then
|` ∩ F| = 1 by 1.8 a), and |` ∩ R2 | = d(A) ≥ 2. Hence, there is a point Q2 ∈ R2 \F.
Subspace operations in affine Klingenberg spaces
105
Since Q2 ∼ R2 , it follows that {P , Q1 , Q2, R1 , S} is independent, and
Q = hP , Q1, Q2i and R = hP , R1 , R2 i
are planes such that Q 6≈ R and Q ∩ R 6= ∅. We note that
F ⊆ Q ∨ R = hP , Q1 , Q2 , R1, R2 i = F ∨ Q2 ⊆ F,
R2 ∈ (Q ∨ R)\F and (A7) yield that Q ∨ R is not free.
Let Q ∈ fm (A) and R ∈ fn (A) such that Q 6≈ R and Q ∩ R 6= ∅, 1 ≤ m ≤ n.
In view of 3.1, our initial problem is to determine when Q ∨ R is free. In fact, as we
shall see, that is the main difficulty in determining a dimension formula for A .
To solve the initial problem, we return to A∗ . From 1.9, Q∗ ∈ fm (A∗ ) and
R∗ ∈ fn (A∗ ). Thus Q∗ ∩ R∗ 6= ∅ and (DF) yield that
dim(Q∗ ∨ R∗) + dim(Q∗ ∩ R∗ ) = n + m.
Of interest now is the relation between (Q∨R)∗ and Q∗ ∨R∗, and the one between
(Q ∩ R)∗ and Q∗ ∩ R∗ . Clearly,
(Q ∩ R)∗ ⊆ Q∗ ∩ R∗ and Q∗ ∨ R∗ ⊆ (Q ∨ R)∗ .
In order to determine these relations, we need to describe Q∗ ∨ R∗. But, as A∗
is also an AK-space, we accomplish this by describing Q ∨ R when Q ∨ R is free.
Thus, in order to determine when Q ∨ R is free, we need to examine the consequences of Q ∨ R being free. We have already one such result.
3.2 ([2], 2.7) Let Q ∈ fm (A) and R ∈ fn (A) such that Q 6≈ R, Q ∩ R 6= ∅ and
Q ∨ R ∈ fn+1 (A), 1 ≤ m ≤ n. Then Q ∩ R ∈ fm−1 (A).
We show that the converse of 3.2 is also valid, and then examine the general
case.
3.3 Proposition. Let {C, F} ⊂ f(A), C 6≈ F. If H = C ∩ F is a hyperplane of
C then C ∨ F ∈ f(A) and F is a hyperplane of C ∨ F.
Proof. Since C 6≈ F , there is a point S0 ∈ C such that S0 6≈ F. If H is a
hyperplane of C then S0 6≈ F and 1.10 c) imply that C = H ∨ S0 . Hence,
F ∨ S0 ⊆ F ∨ C = F ∨ [H ∨ S0 ]
= [F ∨ H] ∨ S0 = F ∨ S0 .
Since S0 6≈ F, F ∨ C = F ∨ S0 is free by 1.5. Clearly, F is a hyperplane of
F ∨ S0 .
3.3.1 Corollary. Let Q ∈ fm (A) and R ∈ fn (A) such that Q 6≈ R and Q ∩ R 6= ∅,
1 ≤ m ≤ n. Then Q ∨ R ∈ fn+1 (A) if and only if Q ∩ R ∈ fm−1 (A).
3.4 Proposition. Let Q ∈ fm (A) and R ∈ fn (A) such that Q 6≈ R and Q∩R 6= ∅,
1 ≤ m ≤ n. Then there is an independent subset {Qt }k1 of Q, 1 ≤ k ≤ m, such that
Q1 6≈ R, Qi 6≈ R ∨ hQt ii−1
for 2 ≤ i ≤ k and R ∨ Q ⊆ R ∨ hQt ik1 ; moreover, if
1
R ∨ Q is free then R ∨ Q = R ∨ hQtik1 and dim(R ∨ Q) = n + k.
Proof. Let R0 ∈ Q ∩ R. Then by (I)n ,there is a basis {Rt }n0 of R. We note that
if such a set {Qt }k1 exists then {Rt }n0 ∪ {Qt }k1 is independent by 1.5. Thus if R ∨ Q
is free then
106
T. Bisztriczky – J.W. Lorimer
hRt in0 ∨ hQt ik1 ⊆ R ∨ Q ⊆ hRt in0 ∨ hQt ik1
and (A7) yield that R ∨ Q = hRt in0 ∨ hQt ik1 .
Next, we determine the existence of {Qt}k1 ⊆ Q. Since Q 6≈ R, there is a
Q1 ∈ Q such that Q1 6≈ R = hRt in0 . By 1.5, {Rt }n0 ∪ {Q1} is independent and
R ∨ Q1 = hRt in0 ∨ Q1 . If Q ⊆ R ∨ Q1 then the assertion is true with k = 1. If
Q 6⊆ R ∨ Q1 then there is a Q2 ∈ Q such that Q2 6≈ R ∨ Q1 = {Rt }n0 ∪ {Q1},
{Rt }n0 ∪ {Qt }21 is independent and R ∨ hQt i21 = hRt in0 ∨ hQt i21 . Again, either
Q ⊆ R ∨ hQti21 or Q 6⊆ R ∨ hQt i21 . Since Q ∈ fm (A), it follows that there is a
smallest k ≤ m such that {R0 } ∪ {Qt}k1 ⊆ Q is independent and {Qt}k1 has the
required property.
3.5 The Dimension Formula for AK-spaces. Let Q ∈ fm (A) and R ∈ fn (A)
such that Q 6≈ R, Q ∩ R 6= ∅ and Q ∨ R ∈ f(A), m ≤ n. Then for some integer k,
1 ≤ k ≤ m,
3.5.1 Q ∨ R ∈ fn+k (A) and Q ∩ R ∈ fm−k (A), and
3.5.2 dim(Q ∨ R) + dim(Q ∩ R) = dim(Q) + dim(R).
Proof. We note that 3.5.2 is an immediate consequence of 3.5.1, and that 3.5.1
has been verified for k = 1 in 3.3.1. Next, 3.4 yields that there is an independent
subset {Qt}k1 ⊆ Q such that Q1 6≈ R, Qi 6≈ R ∨ hQt ii−1
for 2 ≤ i ≤ k and
1
R ∨ Q = R ∨ hQtik1 ∈ fn+k (A). It follows that
Ri : = R ∨ hQt ik−i
∈ fn+k−i (A) for i = 1, ..., k − 1.
1
We set Rk = R and R0 = R ∨ Q. Then Rj+1 is a hyperplane of Rj for
j = 0, ..., k − 1.
Let us consider Q and R1 . Clearly, Q ∩ R1 6= ∅ and from Qk ∈ Q, it follows that
Q 6≈ R1 . We note that
R0 = R ∨ hQt ik1 = R1 ∨ Qk ⊆ R1 ∨ Q ⊆ R0 .
Since Q ∈ fm (A) and R1 is a hyperplane of R1 ∨ Q , it follows by 3.3.1 that
Q1 : = Q ∩ R1 ∈ fm−1 (A) .
Since Qk−1 ∈ Q1 and Q ∩ R ⊆ Q ∩ R2 = Q ∩ R2 ∩ R1 = Q1 ∩ R2 , we obtain
that Q1 ∩ R2 6= ∅, Q1 6≈ R2 and
R1 = R ∨ hQt ik−1
= R2 ∨ Qk−1 ⊆ R2 ∨ Q1 ⊆ R1 .
1
Since Q1 ∈ fm−1 (A) and R2 is a hyperplane of R2 ∨ Q1 , it follows that
Q2 = Q1 ∩ R2 = (Q ∩ R1 ) ∩ R2 = Q ∩ R2 ∈ fm−2 (A) .
Repeating this argument k − 2 more times yields that
Qk = Q ∩ Rk = Q ∩ R ∈ fm−k (A) .
Finally, we present a solution to the problem of when Q ∨ R is free.
Subspace operations in affine Klingenberg spaces
107
3.6 Theorem. Let Q ∈ fm (A) and R ∈ fn (A) such that Q 6≈ R and Q ∩ R 6= ∅ ,
1 ≤ m ≤ n.
3.6.1 If Q ∨ R ∈ f(A) then Q∗ ∨ R∗ = (Q ∨ R)∗ .
3.6.2 Q ∨ R ∈ fn+k (A) if and only if Q ∩ R ∈ fm−k (A) and (Q ∩ R)∗ = Q∗ ∩ R∗,
1 ≤ k ≤ m.
Proof. Let Q ∨ R ∈ f(A). Then by 3.4, Q ∨ R ∈ fn+k (A) for some 1 ≤ k ≤ m
and there is a basis {Rt }n0 ∪ {Qt}k1 of R ∨ Q such that R = hRt in0 and hQt ik1 ⊆ Q.
From 1.4 c), 1.9 and (DF), we obtain that
R∗ ∨ Q∗ ⊆ (R ∨ Q)∗ = h{Rt }n0 ∪ {Qt}k1 i∗
= hR∗t in0 ∨ hQ∗t ik1 ⊆ R∗ ∨ Q∗ ,
Q∗ ∨ R∗ = (Q ∨ R)∗ ∈ fn+k (A∗) and Q∗ ∩ R∗ ∈ fm−k (A∗ ). By 3.5 and 1.9,
(Q ∩ R)∗ ∈ fm−k (A∗). Since (Q ∩ R)∗ ⊆ Q∗ ∩ R∗ and the two spaces are of equal
dimension, (Q ∩ R)∗ = Q∗ ∩ R∗ by (I)m−k .
Conversely, let Q ∩ R ∈ fm−k (A) and (Q ∩ R)∗ = Q∗ ∩ R∗ . Then Q∗ ∩ R∗ ∈
fm−k (A∗ ) and Q∗ ∨ R∗ ∈ fn+k (A∗ ) by 1.9 and (DF). Let {Rt }n0 be a basis of R. By
is a basis of Q ∩ R. We now apply 3.4 to A∗ and
(I)n , we may assume that {Rt }m−k
0
∗
∗
/ R∗ ,
R ∨ Q . Thus there exist an independent subset {Q∗t }k1 ⊆ Q∗ such that Q∗1 ∈
Q∗i ∈
/ R∗ ∨ hQ∗t ii−1
for 2 ≤ i ≤ k and R∗ ∨ Q∗ = R∗ ∨ hQ∗t ik1 .
0
Clearly, we may choose Q1 , ..., Qk in Q. Then Q∗1 ∈
/ R∗ = (hRt in0 )∗ implies that
n
n
Q1 6≈ hRt i0 . Hence {Rt}0 ∪ {Q1 } is independent by 1.5, and h{Rt }n0 ∪ {Q1}i∗ =
/ R∗ ∨ hQ∗t ii−1
yields that {Rt }n0 ∪ {Qt}i1 is
R∗ ∨ Q∗1 by 1.4 c). Similarly, Q∗i ∈
0
independent and h{Rt }n0 ∪ {Qt}i1 i∗ = R∗ ∨ hQ∗t ii1 , 2 ≤ i ≤ k. Since Q ∈ fm (A)
and {Rt }m−k
∪ {Qt}k1 ⊆ Q is independent, it follows that Q = hRt im−k
∨ hQt ik1 by
0
0
(I)m . Finally,
∨ hQt ik1 )
R ∨ hQtik1 ⊆ R ∨ Q = hRt in0 ∨ (hRt im−k
0
= hRt in0 ∨ hQt ik1
= R ∨ hQt ik1
implies that R ∨ Q = hRt in0 ∨ hQt ik1 ∈ fn+k (A).
We gratefully acknowledge that this research was partially supported by the
Natural Sciences and Engineering Research Council of Canada.
108
T. Bisztriczky – J.W. Lorimer
References
[1] T. Bisztriczky and J.W. (Michael) Lorimer, Axiom systems for affine Klingenberg spaces. Research and Lecture Notes in Mathematics, Combinatorics 88, vol.
I (1991), 185-200 Mediterranean Press.
[2] T. Bisztriczky, and J.W. (Michael) Lorimer, On hyperplanes and free subspaces
of affine Klingenberg spaces, Aequationes Mathematicae. 48 No. 2/3 (1994), 121136.
[3] K.W. Gruenberg, and A.J. Weir, Linear Geometry, Springer Verlag, New York,
1977.
[4] A. Kreuzer, An example of a free module in which not every maximal linearly
independent subset is a basis, J. of Geometry, 45 (1992), 105-113.
T. Bisztriczky
Dept. of Mathematics
University of Calgary
Calgary, Canada
T2N 1N4
J.W. Lorimer
Dept. of Mathematics
University of Toronto
Toronto, Canada
M5S 1A1